Question

Help please I will mark brainliest

Answer 1

(a) Fermat's little theorem states that for any prime $p$, we have

$a^{p-1} \equiv 1 \pmod p$

for any integer $a$. Then

$999,999 \equiv 1,000,000 - 1 \equiv 10^6 - 1 \equiv 1 - 1 \equiv 0 \pmod 7$

so 7 does indeed divide 999,999.

(b) Generalizing, we have

$10^{12n} - 1 \equiv \underbrace{999\ldots999}_{12n \text{ nines}} \equiv (10^n)^{12} - 1 \equiv 0 \pmod {13}$

for positive integer $n$. Now, since 1 = 1001 - 1000 and $-1\equiv1000\pmod{1001}$, it follows that $10^3$ is its own inverse modulo 1001, i.e.

$10^3 x \equiv 1\pmod{1001} \implies x = 10^3$

This means

$10^{12n} \equiv (10^3\times10^3)^{2n} \equiv 1^{2n} \equiv 1\pmod{1001} \\\\ \implies 10^{12n} - 1 \equiv 0 \pmod{1001}$

which is what we wanted to show.